Parametric Curves - University of Texas at Austinfussell/courses/cs384g...Algebraic Representation...

University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell

Parametric Curves

University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 2

Parametric Representations3 basic representation strategies:

Explicit: y = mx + bImplicit: ax + by + c = 0Parametric: P = P0 + t (P1 - P0)

Advantages of parametric formsMore degrees of freedomDirectly transformableDimension independentNo infinite slope problemsSeparates dependent and independent variablesInherently boundedEasy to express in vector and matrix formCommon form for many curves and surfaces


Algebraic RepresentationAll of these curves are just parametric algebraic polynomials expressedin different basesParametric linear curve (in E3)

Parametric cubic curve (in E3)

Basis (monomial or power)

!

x = axu3

+ bxu2

+ cxu + dx

y = ayu3

+ byu2

+ cyu + dy

z = azu3

+ bzu2

+ czu + dz

!

x = axu + bx

y = ayu + by

z = azu + bz

!

p(u) = au + b

!

p(u) = au3

+ bu2

+ cu + d

!

u 1[ ]

u3

u2

u 1[ ]


Hermite Curves12 degrees of freedom (4 3-d vector constraints)Specify endpoints and tangent vectors at endpoints

Solving for the coefficients:!

p(0) = d

p(1) = a + b+ c + d

pu(0) = c

pu(1) = 3a + 2b+ c

!

a = 2p(0) " 2p(1) + pu(0) + p

u(1)

b = "3p(0) + 3p(1) " 2pu(0) "p

u(1)

c = pu(0)

d = p(0)

!

pu(u) "

dp

du(u)

•

•pu(0)

u = 0

u = 1

p(0)

p(1)

pu(1)


Hermite Curves - Hermite BasisSubstituting for the coefficients and collecting terms gives

Call

the Hermite blending functions or basis functions

Then

!

p(u) = (2u3" 3u

2+1)p(0) + ("2u

3+ 3u

2)p(1) + (u

3" 2u

2+ u)p

u(0) + (u

3" u

2)p

u(1)

!

H1(u) = (2u

3" 3u

2+1)

H2(u) = ("2u

3+ 3u

2)

H3(u) = (u

3" 2u

2+ u)

H4(u) = (u

3" u

2)

!

p(u) =H1(u)p(0) +H

2(u)p(1) +H

3(u)p

u(0) +H

4(u)p

u(1)

H1 H2

H3

H4

n


Hermite Curves - Matrix FormPutting this in matrix form

MH is called the Hermite characteristic matrixCollecting the Hermite geometriccoefficients into a geometry vector B,we have a matrix formulation forthe Hermite curve p(u)

!

H = H1(u) H

2(u) H

3(u) H

4(u)[ ]

= u3

u2

u 1[ ]

2 "2 1 1

"3 3 "2 "1

0 0 1 0

1 0 0 0

#

$

% % % %

&

'

( ( ( (

=UMH

!

B =

p(0)

p(1)

pu(0)

pu(1)

"

#

$ $ $ $

%

&

' ' ' '

p(u) =UMHB


Hermite and Algebraic Forms

MH transforms geometric coefficients(“coordinates”) from the Hermite basis to thealgebraic coefficients of the monomial basis

!

A =

a

b

c

d

"

#

$ $ $ $

%

&

' ' ' '

p(u) =UA =UMHB

A =MHB

B =MH

(1A

!

MH

"1=

0 0 0 1

1 1 1 1

0 0 1 0

3 2 1 0

#

$

% % % %

&

'

( ( ( (


Cubic Bézier CurvesSpecifying tangent vectors at endpoints isn’t always convenient forgeometric modelingWe may prefer making all the geometric coefficients points, let’s callthem control points, and label them p0, p1, p2, and p3

For cubic curves, we can proceed by letting the tangents at theendpoints for the Hermite curve be defined by a vector between a pairof control points, so that:

!

p(0) = p0

p(1) = p3

pu(0) = k

1(p1"p

0)

pu(1) = k

2(p

3"p

2) p0

p1

p2

• p3p(u)

k2

k1


Cubic Bézier Curves

Substituting this into the Hermite curve expressionand rearranging, we get

In matrix form, this is!

p(u) = 2 " k1( )u3 + 2k

1" 3( )u2 " k1u +1[ ]p0 + k

1u3" 2k

1u2 + k

1u[ ]p1

+ "k2u3 + k

2u2[ ]p2 + k

2" 2( )u3 + 3" k

2( )u2[ ]p3

!

p(u) =UMBP

!

MB

=

2 " k1

k1

"k2

k2" 2

2k1" 3 "2k

1k2

3" k2

"k1

k1

0 0

1 0 0 0

#

$

% % % %

&

'

( ( ( (

!

P =

p0

p1

p2

p3

"

#

$ $ $ $

%

&

' ' ' '


Cubic Bézier CurvesWhat values should we choose for k1 and k2?If we let the control points be evenly spaced in parameterspace, then p0 is at u = 0, p1 at u = 1/3, p2 at u = 2/3 and p3at u = 1. Then

and k1 = k2 = 3, giving a nice symmetric characteristicmatrix:

So

!

MB

=

"1 3 "3 1

3 "6 3 0

"3 3 0 0

1 0 0 0

#

$

% % % %

&

'

( ( ( (

!

p(u) = "u3 + 3u2 " 3u +1( ) p0 + 3u

3" 6u

2 + 3u( )p1 + "3u3 + 3u2( )p2 + u3p3

!

pu(0) = (p

1"p

0) (1/3" 0) = 3(p

1"p

0)

pu(1) = (p

3"p

2) (1" 2 /3) = 3(p

3"p

2)


General Bézier CurvesThis can be rewritten as

Note that the binomial expansion of

(u + (1 - u))n is

This suggests a general formula for Bézier curvesof arbitrary degree

!

p(u) = (1" u)3p0

+ 3u(1" u)2p1

+ 3u2(1" u)p2

+ u3p3

=3

i

#

$ % &

' ( u

i1" u( )

3" ipi

i= 0

3

)

!

n

i

"

# $ %

& ' u

i1( u( )

n( i

i= 0

n

)

!

p(u) =n

i

"

# $ %

& ' u

i1( u( )

n( ipi

i= 0

n

)


General Bézier CurvesThe binomial expansion gives the Bernstein basis (or Bézier blendingfunctions) Bi,n for arbitrary degree Bézier curves

Of particular interest to us (in addition to cubic curves):Linear: p(u) = (1 - u)p0 + up1Quadratic: p(u) = (1 - u)2p0 + 2u(1 - u)p1 + u2p2

!

p(u) =n

i

"

# $ %

& ' u

i1( u( )

n( ipi

i= 0

n

)

Bi,n(u) =

n

i

"

# $ %

& ' u

i1( u( )

n( i

p(u) = Bi,n(u) p

i

i= 0

n

)

Cubic BézierBlending Functions

B0,3

B1,3 B2,3

B3,3n


Bézier Curve PropertiesInterpolates end control points, not middle onesStays inside convex hull of control points

Important for many algorithmsBecause it’s a convexcombination of points,i.e. affine with positive weights

Variation diminishingDoesn’t “wiggle” more than control polygon

p0

p1

p2

p3


Rendering Bézier CurvesWe can obtain a point on a Bézier curve by just evaluatingthe function for a given value of uFastest way, precompute A=MBP once control points areknown, then evaluate p(ui)=[ui

3 ui2 ui 1]A, i = 0,1,2,…,n

for n fixed increments of uFor better numerical stability, take e.g. a quadratic curve(for simplicity) and rewrite

This is just a linear interpolation of two points, each ofwhich was obtained by interpolating a pair of adjacentcontrol points!

p(u) = (1" u)2p0

+ 2u(1" u)p1+ u

2p2

= (1" u)[(1" u)p0

+ up1]+ u[(1" u)p

1+ up

2]


de Casteljau AlgorithmThis hierarchical linear interpolation works for generalBézier curves, as given by the following recurrence

where pi,0 i = 0,1,2,…,n are the control points for adegree n Bézier curve and p0,n = p(u)For efficiency this should not be implemented recursively.Useful for point evaluation in a recursive subdivisionalgorithm to render a curve since it generates the controlpoints for the subdivided curves.

!

pi, j = (1" u)pi, j"1 + upi+1, j"1i = 0,1,2,K,n " j

j =1,2,K,n

# $ %


de Casteljau Algorithm

p0

p1

p2

p3

Starting with the control pointsand a given value of u

In this example, u≈0.25



p0

q0

p1

p2

p3

q2

q1

!

q0(u) = (1" u)p

0+ up

1

q1(u) = (1" u)p

1+ up

2

q2(u) = (1" u)p

2+ up

3



q0

q2

q1

r1

r0

!

r0(u) = (1" u)q

0(u) + uq

1(u)

r1(u) = (1" u)q

1(u) + uq

2(u)



r1p(u)

r0•

!

p(u) = (1" u)r0(u) + ur

1(u)


de Casteljau algorihm

•p0

p1

p2

p3

p(u)


Drawing Bézier Curves

How can you draw a curve?Generally no low-level support for drawing curvesCan only draw line segments or individual pixels

Approximate the curve as a series of line segmentsAnalogous to tessellation of a surfaceMethods:

Sample uniformlySample adaptivelyRecursive Subdivision


Uniform SamplingApproximate curve with n line segments

n chosen in advanceEvaluate

For an arbitrary cubic curve

Connect the points with linesToo few points?

Bad approximation“Curve” is faceted

Too many points?Slow to draw too many line segmentsSegments may draw on top of each other

p4

p0

p1

p2

p3

p(u)

!

pi= p(u

i) where u

i=i

ni = 0,1,...,n

!

pi= a i3 n3( ) + b i2 n2( ) + c i n( ) + d


Adaptive Sampling

Use only as many line segments as you needFewer segments needed where curve is mostly flatMore segments needed where curve bendsNo need to track bends that are smaller than a pixel

Various schemes for sampling,checking results, deciding whetherto sample more

Or, use knowledge of curve structure:Adapt by recursive subdivision

p(u)


Recursive Subdivision

Any cubic curve segment can be expressed as aBézier curveAny piece of a cubic curve is itself a cubic curveTherefore:

Any Bézier curve can be broken up into smaller BéziercurvesBut how…?


de Casteljau constructionpoints are the control pointsof two Bézier sub-segments

xp0

p1

p2

p3

de Casteljau subdivision

q0r0

r1

q2


Adaptive subdivision algorithm

Use de Casteljau construction to split BéziersegmentExamine each half:

If flat enough: draw line segmentElse: recurse

To test if curve is flat enoughOnly need to test if hull is flat enough

Curve is guaranteed to lie within the hulle.g., test how far the handles are from a straightsegment

If it’s about a pixel, the hull is flat


Composite CurvesHermite and Bézier curves generalize line segments to higher degreepolynomials. But what if we want more complicated curves than wecan get with a single one of these? Then we need to build compositecurves, like polylines but curved.Continuity conditions for composite curves

C0 - The curve is continuous, i.e. the endpoints of consecutive curvesegments coincideC1 - The tangent (derivative with respect to the parameter) is continuous,i.e. the tangents match at the common endpoint of consecutive curvesegmentsC2 - The second parametric derivative is continuous, i.e. matches atcommon endpointsG0 - Same as C0

G1 - Derivatives wrt the coordinates are continuous. Weaker than C1, thetangents should point in the same direction, but lengths can differ.G2 - Second derivatives wrt the coordinates are continuous…


Composite Bézier CurvesC0, G0 - Coincident end control pointsC1 - p3 - p2 on first curve equals p1 - p0 on secondG1 - p3 - p2 on first curve proportional to p1 - p0 on secondC2, G2 - More complex, use B-splines to automaticallycontrol continuity across curve segments


Polar form for Bézier CurvesA much more useful point labeling schemeStart with knots, “interesting” values in parameter spaceFor Bézier curves, parameter space is normally [0, 1], and the knotsare at 0 and 1.

Now build a knot vector, a non-decreasing sequence of knot values.For a degree n Bézier curve, the knot vector will have n 0’s followedby n 1’s [0,0,…,0,1,1,…,1]

Cubic Bézier knot vector [0,0,0,1,1,1]Quadratic Bézier knot vector [0,0,1,1]

Polar labels for consecutive control points are sequences of n knotsfrom the vector, incrementing the starting point by 1 each time

Cubic Bézier control points: p0 = p(0,0,0), p1 = p(0,0,1), p2 = p(0,1,1), p3 = p(1,1,1)

Quadratic Bézier control points: p0 = p(0,0), p1 = p(0,1), p2 = p(1,1)

u0 1knot knot


Polar form rulesPolar values are symmetric in their arguments, i.e. allpermutations of a polar label are equivalent.p(0,0,1) = p(0,1,0) = p(1,0,0), etc.Given p(u1, u2,…,un-1, a) and p(u1, u2,…,un-1, b), for anyvalue c we can compute

That is, p(u1, u2,…,un-1, c) is an affine combination ofp(u1, u2,…,un-1, a) and p(u1, u2,…,un-1, b) .

Examples:!

p(u1,u2,...,u

n"1,c) =(b " c)p(u

1,u2,...,u

n"1,a) + (c " a)p(u1,u2,...,u

n"1,b)

b " a

!

p(0,u,1) = (1" u)p(0,0,1) + up(0,1,1)

p(0,u) =(4 " u)p(0,2) + (u " 2)p(0,4)

2

p(1,2,3,u) =(u

2" u)p(2,1,3,u

1) + (u " u

1)p(3,2,1,u

2)

u2" u

1


de Casteljau in polar form

p(0,0,0)

p(1,1,1)

p(0,1,1)

p(0,0,1)



p(0,0,0)

p(1,1,1)

p(0,1,1)

p(0,0,1)

p(0,0,u)p(0,u,1)

p(u,1,1)



p(0,0,0)

p(1,1,1)

p(0,1,1)

p(0,0,1)

p(0,0,u)p(0,u,1)

p(u,1,1)

p(0,u,u)p(u,u,1)



p(0,0,0)

p(1,1,1)

p(0,1,1)

p(0,0,1)

p(0,0,u)p(0,u,1)

p(u,1,1)

p(0,u,u)p(u,u,1)•p(u,u,u)


Composite curves in polar formSuppose we want to glue two cubic Bézier curves togetherin a way that automatically guarantees C2 continuityeverywhere. We can do this easily in polar form.Start with parameter space for the pair of curves

1st curve [0,1], 2nd curve (1,2]

Make a knot vector: [000,1,222]Number control points as before:

p(0,0,0), p(0,0,1), p(0,1,2), p(1,2,2), p(2,2,2)Okay, 5 control points for the two curves, so 3 of themmust be shared since each curve needs 4. That’s whathaving only 1 copy of knot 1 achieves, and that’s whatgives us C2 continuity at the join point at u = 1

u0 1knot knot

u 2knot


de Boor algorithm in polar form

p(0,0,0)

p(0,0,1)p(0,1,2)

p(1,2,2)p(2,2,2)

u = 0.5Knot vector = [0,0,0,1,2,2,2]


Inserting a knot

p(0,0,0)

p(0,0,1)p(0,1,2)

p(1,2,2)p(2,2,2)

u = 0.5

p(0,0,0.5)

p(0,0.5,1)

p(0.5,1,2)

Knot vector = [0,0,0,0.5,1,2,2,2]


Inserting a 2nd knot

p(0,0,0)

p(0,0,1)p(0,1,2)

p(1,2,2)p(2,2,2)

u = 0.5

p(0,0,0.5)

p(0,0.5,1)

p(0.5,1,2)

p(0,0.5,0.5) p(0.5,0.5,1)

Knot vector = [0,0,0,0.5,0.5,1,2,2,2]


Inserting a 3rd knot to get a point

p(0,0,0)

p(0,0,1)p(0,1,2)

p(1,2,2)p(2,2,2)

u = 0.5

p(0,0,0.5)

p(0,0.5,1)

p(0.5,1,2)

p(0,0.5,0.5) p(0.5,0.5,1)

p(0.5,0.5,0.5)

Knot vector = [0,0,0,0.5,0.5,0.5,1,2,2,2]


Recovering the Bézier curves

p(0,0,0)

p(0,0,1)p(0,1,2)

p(1,2,2)p(2,2,2)

Knot vector = [0,0,0,1,1,2,2,2]

p(0,1,1)

p(1,1,2)


Recovering the Bézier curves

p(0,0,0)

p(0,0,1)p(0,1,2)

p(1,2,2)p(2,2,2)

Knot vector = [0,0,0,1,1,1,2,2,2]

p(0,1,1)

p(1,1,2)

p(1,1,1)


B-SplinesB-splines are a generalization of Bézier curves that allows grouping themtogether with continuity across the jointsThe B in B-splines stands for basis, they are based on a very general class ofspline basis functionsSplines is a term referring to composite parametric curves with guaranteedcontinuityThe general form is similar to that of Bézier curves

Given m + 1 values ui in parameter space (these are called knots), a degree n B-splinecurve is given by:

where m ≥ i + n + 1

!

p(u) = Ni,n (u)pi

i= 0

m"n"1

#

Ni,0 (u) =

1 ui$ u < u

i+1

0 otherwise

% & '

Ni,n (u) =

u " ui

ui+n " ui

Ni,n"1(u) +

ui+n+1 " u

ui+n+1 " ui+1

Ni+1,n"1(u)


Uniform periodic basis

Let N(u) be a global basisfunction for our uniformcubic B-splinesN(u) is piecewise cubic

p(u) = N(u) p3+ N(u+1) p2 + N(u+2) p1 + N(u+3)p0

0 4u

N(u)

!

N(u) =

1

6u

3

" 1

2(u "1)

3+ 1

2(u "1)

2+ 1

2(u "1) + 1

6

1

2(u " 2)

3 " (u " 2)2

+ 2

3

" 1

6(u " 3)

3+ 1

2(u " 3)

2 " 1

2(u " 3) + 1

6

=

1

6u

3if u <1

" 1

2u

3+ 2u

2 " 2u + 2

3if u < 2

1

2u

3 " 4u2

+10u " 22

3if u < 3

" 1

6u

3+ 2u

2 " 8u + 32

3otherwise

#

$

% %

&

% %

1 2 3 p0p1p2p3


Uniform periodic B-Spline

p0

p1

p2

p3p(u) = (–1/6u3 + 1/2u2 – 1/2u + 1/6)p0 +

( 1/2u3 – u2 + 2/3)p1 +

(–1/2u3 + 1/2u2 + 1/2u + 1/6)p2 +

( 1/6u3 )p3

Parametric Curves - University of Texas at Austinfussell/courses/cs384g...Algebraic Representation...

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Transcript of Parametric Curves - University of Texas at Austinfussell/courses/cs384g...Algebraic Representation...